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Lecture 5: Additive genetic variance First, a graphical representation of partitioning the variance. Fortunately, variances can be partitioned in to different components. Below VP= total phenotypic variance VG=total genetic variance VE=total environmental variance VGxE=total variance due to the interaction between genotypes and environment. 1
The total genetic variance can be portioned into VA=Additive genetic variance VD=Dominance variation VEPI=Epistatic variation Our main goal today is to understand the partitioning of VA and VD in a one locus, two-­‐allele system (i.e., no possible epistasis). 2
The environmental variance can also be partitioned into different components. 3
h2, heritability in the narrow sense is then: h2 =
VADD
VP H2, heritability in the broad sense is: h2 =
VG
VP The total genetic variance divided by the total phenotypic variance. Only h2 correctly scales the response to selection. Why? 4
Now, what is VA? How can we calculate it from first principles? What are the underlying population genetics? We now take the simplest possible case, and consider one locus, with two alleles, 1 and 2. Hence there is no epistasis. Why? We also assume that there is no environmental variance, VE=0, and as such VGxE is also zero. Why? As previously, p gives the frequency of allele 1. In the drawing above, let a be the phenotype of genotype 11, and -­‐a be the phenotype of genotype 22. Hence the distance between 11 and 22 is 2 times a. (See Falconer & Mackay, 3rd edition.) d is the dominance deviation. In the figure above, allele 1 is partially dominant. d/a is degree of dominance. 5
In the figure below, allele 1 is dominant. Note: our previous dominant coefficient, h = (1 -­‐ d/a)/2. In the figure below, d is “overdominant” (i.e., the heterozygote is more extreme than the homozygotes. 6
Now, we calculate average effects. The average effect (expected deviation from the mean phenotype) of allele 1 is equal to α 1 = p * a + q * d − z Why? Think about the probability of allele 1 finding itself with allele 1, and the probability of finding itself with allele 2. The average effect of allele 2 is equal to α 2 = q *(−a) + p * d − z Now we calculate the breeding values for each genotype. Geno Pheno Breeding value freq 11 a p2 2α 1 12 d 2pq α 1 + α 2 22 -­‐a q2 2α 2 7
The additive genetic variation is equal to the variance in breeding values. In the space below, show how you would calculate the variance in breeding values. Hint see breeding values worksheet on web site. With some algebra (see web site for solution), it can be shown that the variance in breeding values is equal to 2
var(BV) = VA = 2 pq[a + d(q − p)] . (note added: in relating this equation to delta p for pop gen, remember that d is positive if allele 1 is dominate, but negative if allele 1 is recessive.) The dominance variation is equal to 2
VD= (2 pqd) How was that calculated? 8
VA = 2 pq[a + d(q − p)]2 Here are some things to note: 1. the dominance deviation, d, can contribute to VA if q>p. 2. The additive genetic variation can be derived in terms of gene frequencies (p and q), the additive effects of alleles (a), and the dominance deviation (d). Thus there is a population genetic basis to quantitative genetics. 3. The average effects are somewhat abstract quantities, but the breeding values can be measured as 2 times the mean difference between the progeny and the population mean. The mean difference is double because the parent only contributes half the genes. (see Falconer & MacKay, p114) Now, and this is non-­‐trivial, extend your excel worksheet to calculate and graph, h2, VA, VD Due: 15 Sept. Hand in 3 pages (or more) that have the info contained in this example. Show example (add pdf for “PopGen_assign#2_example.pdf”) 9
Curt Lively
W11
W12
W22
s
h
a
d
dominant allele favored: d=1
pheno
1
1
1
1
1 - hs
1
0.8
1-s
-1
0.2 <-- selection coefficient
0 <--equal to (1 - d/a)/2
1 <--Best to leave at 1
1 <--change to scale dominance
1.0
0.9
0.8
0.7
p
herita
V(A)
V(D)
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
gen
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
50
p
0.010
0.012
0.015
0.019
0.024
0.029
0.036
0.044
0.054
0.066
0.080
0.096
0.115
0.137
0.161
0.187
0.215
100
150
Generation
q
0.990
0.988
0.985
0.981
0.976
0.971
0.964
0.956
0.946
0.934
0.920
0.904
0.885
0.863
0.839
0.813
0.785
Wbar
0.804
0.805
0.806
0.808
0.809
0.812
0.814
0.817
0.821
0.826
0.831
0.837
0.843
0.851
0.859
0.868
0.877
pt+1
0.012
0.015
0.019
0.024
0.029
0.036
0.044
0.054
0.066
0.080
0.096
0.115
0.137
0.161
0.187
0.215
0.246
200
250
V(A)
0.0776239
0.0958380
0.1179748
0.1446957
0.1766779
0.2145595
0.2588549
0.3098381
0.3673932
0.4308457
0.4988003
0.5690305
0.6384766
0.7034095
0.7597852
0.8037622
0.8322832
300
V(D)
0.00039204
0.00060353
0.00092579
0.00141388
0.0021477
0.00324101
0.00485206
0.00719464
0.01054686
0.01525346
0.02171514
0.03035786
0.0415764
0.05565268
0.07265964
0.09237377
0.11422737
Curt Lively
W11
W12
W22
s
h
a
d
codominance: d=0
pheno
1
1
1
0.9
1 - hs
0
0.8
1-s
-1
0.2 <-- selection coefficient
0.5 <--equal to (1 - d/a)/2
1 <--Best to leave at 1
0 <--change to scale dominance
1.0
0.9
0.8
0.7
p
herita
V(A)
V(D)
0.6
0.5
0.4
0.3
0.2
0.1
0.0
0
gen
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
50
p
0.010
0.011
0.013
0.014
0.016
0.018
0.020
0.022
0.025
0.028
0.032
0.035
0.040
0.044
0.050
0.056
0.062
100
150
Generation
q
0.990
0.989
0.987
0.986
0.984
0.982
0.980
0.978
0.975
0.972
0.968
0.965
0.960
0.956
0.950
0.944
0.938
Wbar
0.802
0.802
0.803
0.803
0.803
0.804
0.804
0.804
0.805
0.806
0.806
0.807
0.808
0.809
0.810
0.811
0.812
pt+1
0.011
0.013
0.014
0.016
0.018
0.020
0.022
0.025
0.028
0.032
0.035
0.040
0.044
0.050
0.056
0.062
0.069
200
250
V(A)
0.0198000
0.0222164
0.0249196
0.0279416
0.0313173
0.0350847
0.0392853
0.0439637
0.0491677
0.0549484
0.0613599
0.0684584
0.0763024
0.0849514
0.0944648
0.1049006
0.1163135
300
V(D)
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Curt Lively
W11
W12
W22
s
h
a
d
recessive: d = -1
pheno
1
1
1
0.8
1 - hs
-1
0.8
1-s
-1
0.2 <-- selection coefficient
1 <--equal to (1 - d/a)/2
1 <--Best to leave at 1
-1 <--change to scale dominance
1.0
p
herita
V(A)
V(D)
0.8
0.6
0.4
0.2
0.0
0
100
200
300
400
500
600
-0.2
Generation
gen
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
p
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
q
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
0.990
Wbar
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
0.800
pt+1
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
0.010
V(A)
0.0000079
0.0000080
0.0000080
0.0000081
0.0000082
0.0000082
0.0000083
0.0000083
0.0000084
0.0000085
0.0000085
0.0000086
0.0000087
0.0000087
0.0000088
0.0000089
0.0000089
V(D)
0.00039204
0.00039396
0.0003959
0.00039785
0.00039982
0.0004018
0.00040379
0.0004058
0.00040783
0.00040987
0.00041193
0.000414
0.00041608
0.00041818
0.0004203
0.00042244
0.00042459
Why does dominance reduce heritability as the r allele spreads? r=red allele. w=white allele DOMINANCE (r dominant to w) MATING TYPES ww
wr
rr
parent 2à
parent 1
ww
ww---ww
wr
wr---ww
wr---wr
rr
rr---ww
rr---wr
rr---rr
MID-­‐PARENT VALUES (note case for when r allele is rare vs. common) ww
wr
rr
parent 2à
parent 1
ww
white
wr
pink
red
rr
pink
red
red
OFFSPRING ww
wr
rr
parent 2à
parent 1
ww
4W
wr
2W + 2R
1W + 3R
rr
4R
4R
4R
10
CO-­‐DOMINANCE MATING TYPES ww
wr
rr
parent 2à
parent 1
ww
ww---ww
wr
wr---ww
wr---wr
rr
rr---ww
rr---wr
rr---rr
MID-­‐PARENT VALUES (note case for when r allele is rare vs. common) ww
wr
rr
parent 2à
parent 1
ww
white
wr
pink-white
pink
rr
pink
red-pink
red
OFFSPRING ww
wr
rr
parent 2à
parent 1
ww
4W
wr
2pink:2W 1W:2pink:1R
rr
4pink
2R:2pink
4R
11
Note that d is equal to a(1 -­‐ 2h). (where h is our dominance coefficient from pop gen) Substituting, we get VA = 2pq[a – a(1 – 2h)(q – p)]2 Thus if allele 1 is dominant (h = 0) we get VA = 2pqa2[1 -­‐ p + q]2 = 2pqa2[2q]2 = 8pq3a2 if allele 1 is recessive (h = 1), we get VA = 2pqa2[1 + p – q]2 = 2pqa2[2p]2 = 8p3qa2 if allele 1 is co-­‐dominant (h = 1/2), we get VA = 2pqa2 (see “VA(in_terms_of_h).nb” for calculations.) Relate these results to the equation for delta p in lecture 3 on population genetics. a = (W11 – W22)/2 = s/2 thus if allele 1 is dominant (h = 0) we get VA = 8pq3(s/2)2 = 2pq3s2 12
Problem: convert the breeder’s equation into the form used in the QG of correlated traits. 2
!R = h S Δz 1 = h12S 1 !
V
Δz1 = A(1) S1 VP (1)
Δz1 =
VA(1)
VP(1)
The breeder’s equation substitution: R = Δz1 , where the 1 stands for trait 1. VA(1)
substitution: h1 = V P(1)
COV(w, z1 ) substitution: S1 = COV(w, z1 ) COV(w, z1 )
VP(1)
re-­‐arrangement. Δz1 = VA(1)
Δz1 = VA(1) B(w, z1 ) substitution: COV(w, z1 )
= B(w, z1 ) , the selection gradient. VP(1)
13
Δz1 = G11B1 Δz = GB ⎡ Δz1 ⎤ ⎡ G11 G12
⎢ ⎥ ⎢
⎢ Δz2 ⎥ ⎢ G21 G22
⎢ ⎥ ⎢
.
⎢ . ⎥ ⎢ .
⎢ . ⎥ =⎢ .
.
⎢ ⎥ ⎢
⎢ . ⎥ ⎢ .
.
⎢ ⎥ ⎢
⎢⎣Δzp ⎥⎦ ⎢⎣ Gp1 Gp2
VA(1) = G11 and B(w,z1) =B1 the standard form for correlated traits, where G is the additive genetic variance/covariance matrix, and B is the vector of selection gradients .
.
.
.
.
.
.
.
.
G1p ⎤ ⎡ B1 ⎤
⎥⎢ ⎥
G2 p ⎥ ⎢ B2 ⎥
⎥⎢ ⎥
. ⎥⎢ . ⎥
. ⎥⎥ ⎢⎢ . ⎥⎥
. ⎥⎢ . ⎥
⎥⎢ ⎥
Gpp ⎥⎦ ⎢⎣ Bp ⎥⎦
the matrix/vector expansion Thus, for two traits, we get 14
Δz1 = G11B1 + G12 B2
and
Δz2 = G21 B1 + G22 B2
where G11 = the additive genetic variance for trait 1, and G22 = the additive genetic variance for trait 2. And: G12= G21= Additive genetic covariance between traits 1 & 2. Now, the selection gradients here are not simple regression coefficients as in the single-­‐trait case. The selection gradients are coefficients in a multiple regression. S 1VP '2) − S 2COV' z 1 , z 2 )
B1 =
expansion of a partial regression coefficient VP '1)VP '2) − COV' z 1 , z 2 )2 !
Note that if the COV=0, then… 15
Note that if the COV=0, then… S 1VP '2) − 0
S
B1 =
= 1
VP '1)VP '2) − 0 VP '1) !
and… S 1 V A)1*
Δz 1 = G11B1 = G11
=
S 1 = h 2S 1
V p )1* V p )1*
!
But if the covariance is not equal to zero, selection on trait 2 affects the selection gradient for trait 1. This makes intuitive sense. Yes? Here is the main point. Selection on a trait can affect other traits. This could affect the magnitude and even the direction of change. See article on web by Steve Arnold. For derivation of partial regression coefficients, see page 183: Evolutionary Quantitative Genetics by Derek A. Roff. Note, equation 5.13 in Roff’s book has a typo: for Bx the VPX in the numerator should be VPY. And for BY the VPY in the numerator should be VPX. 16